Difference between revisions of "2006 AIME II Problems/Problem 7"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
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There are two other scenarios. <math>a</math> and <math>b</math> can be a three digit number and a two digit number, or a three digit number and a one digit number. For the first scenario, there are <math>9 \times 8 \times 2=144</math> possibilities (the two accounting for whether <math>a</math> or <math>b</math> has three digits) and for the second case there are <math>9 \times 2=18</math> possibilities. Thus, thus total possibilities for <math>(a,b)</math> is <math>576+144+18=738</math>. | There are two other scenarios. <math>a</math> and <math>b</math> can be a three digit number and a two digit number, or a three digit number and a one digit number. For the first scenario, there are <math>9 \times 8 \times 2=144</math> possibilities (the two accounting for whether <math>a</math> or <math>b</math> has three digits) and for the second case there are <math>9 \times 2=18</math> possibilities. Thus, thus total possibilities for <math>(a,b)</math> is <math>576+144+18=738</math>. | ||
− | |||
=== Solution 3 === | === Solution 3 === | ||
We first must notice that we can find all the possible values of <math>a</math> between <math>1</math> and <math>500</math> and then double that result. | We first must notice that we can find all the possible values of <math>a</math> between <math>1</math> and <math>500</math> and then double that result. | ||
− | When <math>1</math> < <math>a</math> < <math> | + | When <math>1 < a < 100</math> there are <math>9\times9 = 81</math> possible solution for <math>a</math> so that neither <math>a</math> nor <math>b</math> has a zero in it, counting <math>1</math> through <math>9</math>, <math>11</math> through <math>19</math>, ..., <math>81</math> through <math>89</math>. |
− | + | When <math>100 < a < 200</math> there are <math>9\times8 =72</math> possible solution for a so that neither a nor b has a zero in it, counting <math>111</math> through <math>119</math>, <math>121</math> through <math>129</math>, ..., <math>181</math> through <math>189</math>. | |
− | This can | + | This can clearly be extended to <math>100k < a < 100(k+1)</math> where <math>k</math> is an integer and <math>0 <k < 9</math>. |
− | + | Thus for <math>100 < a < 500</math> there are <math>72\times4</math> = <math>288</math> possible values of <math>a</math>. | |
+ | |||
+ | Thus when <math>1 < a < 500</math> there are <math>288 + 81 =369</math> possible values of <math>a</math> and <math>b</math>. | ||
+ | |||
+ | Doubling this yields <math>369\times2= 738</math>. | ||
+ | |||
+ | === Solution 4 (Similar to Solution 2)=== | ||
+ | We proceed by casework on the number of digits of <math>a.</math> | ||
+ | |||
+ | Case 1: Both <math>a</math> and <math>b</math> have three digits | ||
+ | |||
+ | We now use constructive counting. For the hundreds digit of <math>a,</math> we see that there are <math>8</math> options - the numbers <math>1</math> through <math>8.</math> (If <math>a = 9,</math> that means that <math>b</math> will be a two digit number, and if <math>a = 0,</math> <math>a</math> will have two digits). Similarly, the tens digit can be <math>1-8</math> as well because a tens digit of <math>0</math> is obviously prohibited and a tens digit of <math>9</math> will lead to a tens digit of <math>0</math> in the other number. The units digit can be anything from <math>1-9.</math> Hence, there are <math>8 \cdot 8 \cdot 9 = 576</math> possible values in this case. | ||
+ | |||
+ | Case 2: <math>a</math> (or <math>b</math>) has two digits | ||
+ | |||
+ | If <math>a</math> has two digits, the only restrictions are that the units digit must not be <math>0</math> and the tens digit must not be <math>9</math> (because then that would lead to <math>b</math> beginning with <math>90...</math>). There thus are <math>8 \cdot 9 = 72</math> possibilities for <math>a,</math> and we have to multiply by <math>2</math> because there are the same number of possibilities for <math>b.</math> Thus, there are <math>72 \cdot 2 = 144</math> possible values in this case. | ||
+ | |||
+ | Case 3: <math>a</math> (or <math>b</math>) has one digit | ||
+ | |||
+ | This is easy -- <math>a</math> can be anything from <math>1</math> to <math>9,</math> for a total of <math>9</math> possible values. We multiply this by <math>2</math> to account for the single digit <math>b</math> values, so we have <math>9 \cdot 2 = 18</math> possible values for this case. | ||
+ | |||
+ | Adding them all up, we get <math>576 + 144 + 18 = \boxed{738},</math> and we're done. | ||
+ | |||
+ | Solution by Ilikeapos | ||
+ | |||
+ | === Solution 5=== | ||
+ | For every <math>a \in [1, 999]</math>, <math>(a, 1000 - a)</math> is a potential candidate for a solution, barring the cases where <math>a</math> or <math>1000 -a </math> has zero digits. | ||
+ | |||
+ | First, let's consider all viable <math>a</math> that do not have a zero digit. As there are 9 non-zero digits, we have: | ||
+ | * <math>9^1</math> 1-digit numbers without a zero digit | ||
+ | * <math>9^2</math> 2-digit numbers without a zero digit | ||
+ | * <math>9^3</math> 3-digit numbers without a zero digit | ||
+ | |||
+ | However, we have still overlooked the cases where <math>1000 - a</math> contains zero digits: | ||
+ | * '''Case 1''': If the one's digit of <math>1000 - a</math> is zero, then <math>a</math> will trivially end in a zero, which we've already excluded. | ||
+ | * '''Case 2''': If the ten's digit of <math>1000 - a</math> is zero, with digital representation <math>\overline{X0Y}</math>, then <math>a</math> has the digital representation <math>\overline{[9-X]9[10-Y]}</math>. <math>X</math> and <math>Y</math> can each take on any value in <math>[1, 9]</math> to produce a value in our set of potential <math>a</math>. There are thus <math>9^2</math> cases that we overlooked, where <math>a</math> had no zero digits, but <math>1000 - a</math> did. | ||
− | + | Adding up the cases with <math>a \in [1, 999]</math> with no zero digits and removing the cases with <math>1000 - a</math> with zero digits gives us <cmath>(9^1 + 9^2 + 9^3) - 9^2 = \boxed{738}</cmath> | |
− | + | ~SaifHakim | |
== See also == | == See also == | ||
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[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 03:41, 12 September 2021
Problem
Find the number of ordered pairs of positive integers such that and neither nor has a zero digit.
Contents
Solution
Solution 1
There are numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when or have a 0 in the tens digit, and since the equation is symmetric, we will just count when has a 0 in the tens digit and multiply by 2 (notice that the only time both and can have a 0 in the tens digit is when they are divisible by 100, which falls into the above category, so we do not have to worry about overcounting).
Excluding the numbers divisible by 100, which were counted already, there are numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling such numbers; considering also and we have . Therefore, there are such ordered pairs.
Solution 2
Let and be 3 digit numbers:
cde +fgh ---- 1000
and must add up to , and must add up to , and and must add up to . Since none of the digits can be 0, there are possibilites if both numbers are three digits.
There are two other scenarios. and can be a three digit number and a two digit number, or a three digit number and a one digit number. For the first scenario, there are possibilities (the two accounting for whether or has three digits) and for the second case there are possibilities. Thus, thus total possibilities for is .
Solution 3
We first must notice that we can find all the possible values of between and and then double that result.
When there are possible solution for so that neither nor has a zero in it, counting through , through , ..., through . When there are possible solution for a so that neither a nor b has a zero in it, counting through , through , ..., through . This can clearly be extended to where is an integer and . Thus for there are = possible values of .
Thus when there are possible values of and .
Doubling this yields .
Solution 4 (Similar to Solution 2)
We proceed by casework on the number of digits of
Case 1: Both and have three digits
We now use constructive counting. For the hundreds digit of we see that there are options - the numbers through (If that means that will be a two digit number, and if will have two digits). Similarly, the tens digit can be as well because a tens digit of is obviously prohibited and a tens digit of will lead to a tens digit of in the other number. The units digit can be anything from Hence, there are possible values in this case.
Case 2: (or ) has two digits
If has two digits, the only restrictions are that the units digit must not be and the tens digit must not be (because then that would lead to beginning with ). There thus are possibilities for and we have to multiply by because there are the same number of possibilities for Thus, there are possible values in this case.
Case 3: (or ) has one digit
This is easy -- can be anything from to for a total of possible values. We multiply this by to account for the single digit values, so we have possible values for this case.
Adding them all up, we get and we're done.
Solution by Ilikeapos
Solution 5
For every , is a potential candidate for a solution, barring the cases where or has zero digits.
First, let's consider all viable that do not have a zero digit. As there are 9 non-zero digits, we have:
- 1-digit numbers without a zero digit
- 2-digit numbers without a zero digit
- 3-digit numbers without a zero digit
However, we have still overlooked the cases where contains zero digits:
- Case 1: If the one's digit of is zero, then will trivially end in a zero, which we've already excluded.
- Case 2: If the ten's digit of is zero, with digital representation , then has the digital representation . and can each take on any value in to produce a value in our set of potential . There are thus cases that we overlooked, where had no zero digits, but did.
Adding up the cases with with no zero digits and removing the cases with with zero digits gives us
~SaifHakim
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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